 While solving Cauchy problem for quasi-linear equations, one of the steps was to pass characteristic curves through points of the datum curve. For that, what we do is we solve a system of characteristic ODE with initial condition so that the point initial condition point lies on the datum curve. Now when it comes to the general non-linear equations, if you want to do the same thing, we did not have a complete system for the characteristic ODE because it involved p and q which were unknown. Therefore, we extended the system to a system of characteristic strip, system of differential equations for a characteristic strip. Now we would like to solve this but now the problem is that characteristic strip is not known on the datum curve. On the datum curve only the values or the initial conditions for x, y, z will be known. Therefore, we need to derive initial strip that is what we are going to do in this lecture. We start with a recap of what we have done so far in the characteristic method for general non-linear equations and then we find the initial strip and then we take the third step which is to define a candidate solution and first steps in the step 3. In the next lecture we are going to conclude the entire all the four steps. So this is to just make you recall the notation ql stands for quasi-linear equations AUX plus BUI equal to C, GE we call it general non-linear equations sometimes people call fully non-linear equations it is f of x, y, u, x, y equal to 0. Now in the solution of the Cauchy problem for general non-linear equations where are we let us look at that. Recall the key steps involving in the solution of Cauchy problem that we proposed for general non-linear equations. Step 1 is obtaining a system of ordinary differential equations for characteristic strip which we have done. Step 2 is finding an initial strip we are going to do today and step 3 defining a candidate solution and establishing that the candidate solution is indeed a solution to Cauchy problem will be taken up in the next lecture. So step 1 was successfully implemented so far. Let us recall the difficulties which are new to GE when compared to ql and whatever the ideas that helped us to overcome them. Difficulties and the resolution in step 1 our analysis was motivated by the quasi-linear equations. Quasi-linear equation ql gave us a characteristic direction. Characteristic direction gave rise to characteristic curves, characteristic curves made up an integral surface. The difficulty 1 for general non-linear equation is the equation does not give away a characteristic direction. A useful idea was we observed that for quasi-linear equations the characteristic direction is the envelope of possible tangent planes. Actually envelope of possible tangent planes is a straight line whose direction is a characteristic direction. So we found that the same idea works for GE as well. A characteristic direction at a point p xyz is given by fp fq pfp plus qfq fpfq denote partial derivatives of f with respect to the variable p and q respectively. And this small pq satisfy f of xyz pq equal to 0 capital P actually is standing for xyz. Now what is the difficulty number 2? Characteristic OD system is incomplete for GE because we were looking at finding a curve having characteristic direction which was given by this dx by dt dv by dt dz by dt equal to fp fq pfp plus qfq. We were looking at solutions to this whose image will be a curve. And with this condition x0 equal to x0 y0 equal to y0 z0 equal to z0. Then this will be a characteristic curve passing through the point x0 y0 z0. Unfortunately in this system pq are involved and pq also depend on the location where you are on the characteristic curve that you are trying to find. So p and q are dependent on xt yt zt. That is why Kara OD that is this system is called Kara OD. It is not soluble. So we need to complete this system that is what we did. We derived a system of OD is for the characteristic strip which is a system of 5 equations. Now we need to find a solution to this. What is the next step? As I told you at the beginning of this lecture is to find solutions to this so that x0 y0 z0 is a point on the datum curve let us say fs gshs. But what will be the initial conditions for p and q? Those are not there. So we need to derive that. That is what is called finding an initial strip. Initial curve that is a datum curve is known fs gshs as s varies in the interval i we are going throughout the curve gamma. Now we need to find ps and qs and ps and qs are not arbitrary functions. They should be such that ps qs minus 1 should be the normal to sorry to a possible tangent plane that we want to find. So it is something tied with the equation the ps qs. For this step we are going to assume that the initial data is c2 fgh are all c2 functions. This is only a temporary requirement. We will see a comment later on saying that okay this is not needed at some point. As I told you to derive the equations we can assume anything that we want. But having got the equations we have to try to show that things work with the minimal assumptions. Assuming fgh is c1 is somewhat reasonable c2 is too much. It is not that much reasonable. But this we are going to do only to derive this an initial strip okay. We will see that. We will also see why or where are we going to need this c2 s. So finding an initial strip. So step 2 in solving Cauchy problem is to find an initial value problem for Kara strip ODE. The datum curve gives rise to the initial conditions for xyz. We need to find initial conditions for p and q. In other words we need to extend the datum curve to an initial strip. The support of the strip will be the datum curve. So finding an initial strip how do we do that. So let zeta t be xt by tz t ptqt be a solution to Kara strip ODE. So it follows from Kara strip ODE that along a characteristic strip d by dt of this is 0. How does it follow? You do the chain rule. So it will be fx into x dash t plus fy into y dash t, fz into z dash t, fp into p dash t plus fq into q dash t. And if you use these equations it turns out that you end up with 0. Therefore d by dt of f of xt by tz t ptqt is 0. So that means this function t going to f of zeta t is a constant function because its derivative is 0 it has to be constant. If it is constant it will be equal to f of zeta of t0 for any t0 but I will choose t equal to 0 because t equal to 0 is where I am going to stay on the datum curve at least x0, y0, z0 will be on the datum curve. Therefore we are going to require that f of zeta 0 is 0 that is what we are going to ask. So if we choose an initial strip zeta 0 which is x0, y0, z0, p0, q0 such that f of that is 0 it means f of zeta t will be 0 for all t in j. So recall that the Cauchy data is given by x equal to fs, y equal to gs, z equal to hs. We are interested in passing a characteristic curve through every point of gamma. Since karastrip ODE is a coupled system involving p and q also apart from x, y, z which only matters to find a characteristic curve to find a characteristic curve you need xt by tz t but the equations involve p and q. So if we first have to determine an initial strip having the datum curve as its support. In other words at every point of the datum curve if this is gamma any point we need to associate a vector I am saying vector because two numbers are there and this should be having this finally this property that this is ps the point is s this is fs gs hs. The strip is fs gs hs ps qs the support is fs gs hs. Since the integral surface must contain a part of gamma on it we should have hs equal to u of fs gs it should hold for s belonging to a sub interval of i. Now once you have that differentiate this equation with respect to s you get h dash s equal to ux into f dash plus uy into g dash by chain rule. Since ps equal to p is supposed to be ux right q is supposed to be uy that should also hold so we demand this ps equal to this should happen qx equal to this should happen therefore what does this imply. The equation that we obtained in the previous slide h prime equal to ux f prime uy g prime now ux fs gs must be p and uy must be q therefore we need to solve for ps and qs in terms of s coming from this equation capital F of fs gs hs comma p q equal to 0 and pf dash s plus q g dash s equal to h dash s. So phi b is this equation and this is the equation of the partial differential equation. The role p and q are actually for ux and uy so that is why we demand these two conditions. So now if you notice this is a function of fs gs hs are given therefore f dash is known g dash is known h dash is known. So what it involves is p and q and a function of s something s. So you can think of this as a function of s pq equal to 0 another function of s pq equal to 0 and you want to solve pq in terms of s. So implicit function theorem says that if we know a special solution for some s equal to s not p not q not which satisfies this as well as this that means capital F of f of s not comma g of s not comma h of s not comma p not q not equal to 0 and p not f dash s not plus q not g dash s not equal to h dash of s not if such p not q not are given and a certain Jacobian condition is satisfied non-zeroness of certain Jacobian we will do the details on the next slides. Then there will exist functions you can express pq in terms of s essentially that is what we want. So there exist functions s going to ps s going to qx of course the conclusions are local therefore these functions will be defined on an interval containing s0 where this Jacobian condition is satisfied and where a particular solution p not q not has been found such that these functions are differentiable and they are solutions to the above system. Moreover they are unique with respect to the above properties implicit function theorem when applicable it gives a unique solution. So when s equal to s not pq equal to p not q not if that is a solution to the system of equations that is exactly this which I have read out earlier. So in order to put our problem in the setup of implicit function theorem we define two functions p1 and p2 as I told you p1 of s comma pq because fs gshs are known it is simply a function of s that is why s pq are unknown quantities I want to solve pq in terms of s. So I have 3 variables here so s varies in interval i pq vary in the interval r cross r I have put but it should be that wherever capital F is defined for this pq right. So it will be the projection to the last two coordinates pq of the domain of omega 5 that is where it it makes sense. So if you assume that that is r omega 5 is such that the last two components r then is r cross r otherwise it is going to be the corresponding projections to the last two coordinates that will come in place of r cross r because f should be meaningful after all right here I am defining this function for p and q belonging to r and r of course that makes sense only if f is defined for those p and q's. So set up all pq for which f is defined I will consider this function in that case I will not write r cross r I will write something else okay. So with this correction p2 of spq is the second equation that we had on the previous slide we are interested in its solution what is given is s naught p naught q naught is a solution you have to find the p naught q naught such that s naught p naught q naught is a solution. So that is what we assumed now to apply implicit function theorem we need to check that functions are c1 functions phi 1 phi 2 are c1 functions are they yeah with respect to s if fgh or c1 fine this is c1 and capital F itself is c1 therefore composition will give c1 pq are appearing in this and we already assumed f is c1 with respect to all the components therefore pq also will be c1. So phi 1 will be c1 no problem phi 2 c1 s of phi 2 will involve c2 s of f so this is where we need fgh to be c2 okay. In terms of p and q it is linear so it is c infinity so phi 2 is c1 provided we assume the initial data that is datum curve fgh or c2 functions this is where we need fgh to be c2 okay fine. Now the Jacobian of phi 1 phi 2 with respect to pq at the solution that we already found s naught p naught q naught of this system that should be non-zero which is this dou phi 1 phi 2 by dou pq is simply the Jacobian which is here dou phi 1 by dou p dou phi 1 by dou q dou phi 2 by dou p dou q will come now at the point s naught p naught q naught that you can compute from the expressions of phi 1 phi 2 it will turn out to be fp at zeta naught fq at zeta naught f prime s naught g prime zeta naught is this phi tuple fs naught gs naught hs naught p naught q naught this is non-zero we have to assume that assume the above condition on Jacobian okay done now we can apply implicit function theorem it gives us unique functions p equal to ps and q equal to qs which are c1 on an interval containing s equal to s naught let us still denote it by i for convenience if you want you can make it i dash does not make any difference and what is the property ps and qs solve a system phi 1 and phi 2 phi 1 s phi s equal s ps qs equal to 0 phi 2 of s ps qs equal to 0 and at s equal to s naught it coincides with p naught q coincides with q naught thus an initial strip has been determined so this completes successful implementation of step 2 of our program so initial strip consists of the datum curve gamma which is given to us and along with planes planes essentially means the normal to the planes ps qs minus 1 attached at each of the points of gamma okay what we have shown is that if the Jacobian condition is met at an s equal to s naught in a piece nearby that is some part of gamma there we can do this that is what the implicit function theorem said in any case our theorem is going to be local we are always going to fix some point and going to assert that near this point p0 on the datum curve gamma there is an integral surface which contains a piece of gamma therefore this is absolutely fine and initial strips have the datum curve as the support okay if you can find a throughout all the points okay now let us go to the step 3 which is defining a candidate solution to define a candidate solution first thing is we have to solve characteristic strip equation with the initial strip that we have got so here we assume that f is c2 and so let this represent solutions to characteristic ODE satisfying the initial conditions for x, y, z it is fs gs hs for p and q it is small p and small q of functions of s which we have determined in step 2 that is a initial strip that means if you look at the trajectory of the solutions the first three coordinates at t equal to 0 pass through the point fs gs hs on the datum curve and remaining 2 are supposed to be giving you the components of the normal at that point that is why the strip now existence of solutions like this follows from Cauchy-Lipschitz Picard theorem that is why I assumed f is c2 because the right hand side will involve fp fq etc right so to guarantee that right hand side is Lipschitz one easy way to do is assume that right hand side is c1 then it will be locally Lipschitz and then we have existence of unique solutions so that would require f is c2 that is why I have added f is c2 so solutions are unique because of that otherwise you have existence continuity is enough we assumed f is c1 and if you look at the equations of characteristic ODE the right hand side will involve only derivatives of f at maximum right fp fq pfp plus pfq minus fx fz and so on f y they are all assumed to be continuous therefore right hand side is continuous the moment you assume f is c1 which means by Pino's theorem there exists a solution if you want uniqueness you push it c2 if you make a f to be c2 uniqueness is guaranteed so on solutions of the initial value problem that is characteristic strip ODE plus an initial strip we want to solve this initial value problem so without loss of generality we may assume that solutions to the initial value problem which is given by the ODE is a characteristic strip ODE and the initial conditions are given by the initial strip they are defined the solutions are defined for t, s in j cross i where j does not depend on s for quasi linear equations it was possible thanks to the lemma on reparameterization of characteristic curves because the tangential direction does not change under reparameterization on the other hand reparameterization of a characteristic strip need not be a characteristic strip therefore this argument does not hold luckily we have another argument in the case of quasi linear equations we had a second argument which we have given that also allows us to conclude the existence of such an interval j which does not depend on s but now there is a compromise that the interval i needs to be replaced with a sub interval of y but that does not matter because later on what we are going to do is we are going to apply inverse function theorem whose conclusions are anyway local so it does not matter so what is important is that j does not depend on s whether it is s belongs to i or s belongs to some sub interval of i so the solutions are continuously differentiable functions on j cross i by differentiable dependence of solutions and parameters in the theory of ODE's in fact inside the proof of this theorem one needs to get such catch hold of such an interval which does not depend on s so the progress so far has been this first gamma which is given then an initial strip we have obtained that is throughout these points of gamma we have erected these kind of small planes and then characteristic curves going through points of gamma if this is the point p this is the gamma p small gamma p and not only characteristic curves throughout every at any point in the characteristic curve we also got this particular normal to some plane of course the plane is expected to be time in plane later on so for T s in j cross i we got f of x T s y T s z T s p T s q T s equal to 0 and x 0 s is f s y 0 s is g s z 0 s is h s p is p s q is q s so these are coming from initial conditions this is the first thing that we did d by dt f of zeta of t is 0 therefore f of zeta t is constant and we chose zeta 0 such that f of zeta 0 is 0 therefore f of zeta t is 0 so this is also we have proved in step 2 it followed from the way we chose initial strip so summarizing the steps 1 2 and the third one the little bit that we did so far it gives us the following theorem consider the Cauchy problem for general nonlinear equation with Cauchy data assumptions f is c 2 let s be an integral surface for this general nonlinear equation where u is a c 2 function assume that a part of the datum curve lies on the surface yes assume that f g h are c 2 functions which describe the datum curve note these are additional assumptions on f and gamma standard assumptions on them still apply namely f p f q cannot simultaneously vanish and so on and f prime g prime also cannot vanish simultaneously now take a point which is on gamma as well as on s that means that integral surface which is also there in the datum curve that point let p naught q naught be such that the equations are satisfied assume that the Jacobian condition is satisfied at s naught p naught q naught p naught q naught is this s naught is which we have already fixed and here the zeta naught stands for this as before conclusion the surface is the union of the supports of the characteristic strips through the points of gamma of course in a neighborhood of p 0 so this theorem actually generalizes our earlier theorem that we did integral surface is a union of characteristic surface curves integral surface is union of characteristic curves that theorem we proved for Cauchy equations this is generalization of that proof is similar to that same as that so key ingredients here are characteristic ODE and p t equal to this and q t equal to this so now question is the road ahead is smooth because last theorem gives us a hope to find an integral surface exactly like the other theorem which gave us hope in quasi linear equations using the method of characteristics for general and non-linear equations however theorem holds only for u in C2 and Fgh in C2 of i we are looking to solve GE for which a C2 solution may not exist it is unreasonable to expect a C2 solution for a first order PDE to start with it may happen that you you have a C2 solution or C infinity solution but to start with you do not expect that and parameterization of the datum curve is C1 is reasonable but not C2 so would that be a problem answer is no that would not be a problem details on the next slide yes it is true that we used u is C2 of D Fgh is in C2 of i to derive characteristic ODE and initial strip now forget all this and start working with characteristic ODE and initial strip they ever used only to derive these two things right pretend that you are given these two and you are asked to work with them everything will be fine we do not need Fgh in C2 so no need to worry road ahead is indeed smooth from now onwards assume that the system characteristic ODE is given and an initial strip FSGS HSPSQS consisting of C1 functions on an interval containing S0 is given let the functions HTS, YTS, ZTS, PTS, QTS solve the initial value problem and initial strip that will be good enough of course you may not have solutions which are unique but may be correspond to each solution you may get one integral surface who knows that that can happen so we derived an initial strip using which the system of characteristics strip equations may be solved projection of solutions to XYZ space will give you characteristic curves through points of gamma in a bit to extend method of characteristics to GE we come across new geometrical entities so natural question you should be still call it method of characteristics or should we call method of characteristic strips it is okay that does not matter just for you to think ponder about this question so we found an initial strip step 2 of our program is successfully implemented in the next lecture we are going to complete the proof of existence and uniqueness of solutions to Cauchy problem for general nonlinear equations thank you